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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Sep 2019, 13:59
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ? A. 142 B. 71 C. 99 D. 20 E. 18 PS01661.01
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Sep 2019, 17:07
The difference between the two digits (abc − cba) = \(100a+10b+c  100c10ba = 99a99c = 99(ac)\) As 99 is not divisible by 7, then \((ac)\) must be divisible by 7 (and b has no effect on the overall outcome)
\((ac)\) are only 2 possible values: \((92)\) and \((81)\) ;as a,b,c are non zero; and a must be > c to keep the positive sign. b has \(9\) possible values (all except 0)
So the total number of possible outcomes = \(2*9 = 18\) E




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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Sep 2019, 22:38
gmatt1476 wrote: The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142 B. 71 C. 99 D. 20 E. 18
PS01661.01 The numbers are 100a + 10b +c and 100c + 10b + a. After taking the difference, we are left with 99(ac). Now for this number to be a multiple of 7, ac = 7 or 0. for (ac) = 7, there are two possibilities > (9,2) and (8,1) and for ac = 0, a = c, so there are 9 possibilities that are 1 to 9. The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case. There are a total of 11 cases so the total number of numbers possible are 9*11 = 99. IMO, answer is C



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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22 Sep 2019, 01:32
given 100a+10b+c100c10ba ; 99(ac) for 99(ac) to be divisible (ac) has to be divisible by 7 and a,b,c are all single digit non zero ; only possibility of (ac) = 7 ; ( 92) & (81) ; 2 choices of a & c ; for b which is middle term we have 9 options ; so total possible values; 9*2 ; 18 IMO E gmatt1476 wrote: The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142 B. 71 C. 99 D. 20 E. 18
PS01661.01



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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26 Sep 2019, 08:58
gmatt1476 wrote: The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142 B. 71 C. 99 D. 20 E. 18
PS01661.01 Given: The difference 942 − 249 is a positive multiple of 7. Asked: If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ? (100a + 10b + c)  (100c + 10b + a) = 99(ac) Since 99 is not a multiple of 7, (ac) should be a multiple of 7. There are 2 possibilities = {(8,1),(9,2)} There are 9 possibilities for b Total such numbers = 2*9 = 18 IMO E



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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26 Sep 2019, 22:10
One question on this  If a, b and c are given as different alphabets, should we consider them as distinct or not?



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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26 Sep 2019, 22:24
Kanika3agg wrote: One question on this  If a, b and c are given as different alphabets, should we consider them as distinct or not? No, You can't consider them distinct without being given that they are different/distinct integers or something such as a≠b≠c or a>b>c



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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26 Sep 2019, 22:56
MahmoudFawzy  Thank you!



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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11 Oct 2019, 07:28
Brilliant question. Can someone please post questions like this one ?



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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04 Nov 2019, 23:32
uchihaitachi wrote: gmatt1476 wrote: The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142 B. 71 C. 99 D. 20 E. 18
PS01661.01 The numbers are 100a + 10b +c and 100c + 10b + a. After taking the difference, we are left with 99(ac). Now for this number to be a multiple of 7, ac = 7 or 0. for (ac) = 7, there are two possibilities > (9,2) and (8,1) and for ac = 0, a = c, so there are 9 possibilities that are 1 to 9. The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case. There are a total of 11 cases so the total number of numbers possible are 9*11 = 99. IMO, answer is CBy this method, the number 252, 343, 515 etc. all qualify to represent 'abc'. Let's take 252 as abc. cba = 252. abc  cba = 0. But the question wants the difference to be a positive multiple of 7. Now, 0 is neither positive nor negative.Thus, when abc = 252, then abc – cba is not a positive multiple of 7. The condition a = c is not valid.Only (a,c) = (9,2) and (8,1) is possible. In both cases b can take 9 possible values as b ≠ 0. 2 x 9 = 18 possible choices.



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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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14 Dec 2019, 19:11
Can we solve the problem this way?
942 = 3^2 2^2 2^1
Total Factors = (2+1) * (2+1) * (1+1) = 18



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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Jan 2020, 02:57
Notice something cool in this question, for abc  cba to be divisible by 7, ac must equal 7 (you can see that from the question stem) if you don't have a strong number sense, using algebra you could reach the same conclusion.
(1) a,c,b are nonzero digits and they are also different (2) ac will equal 7 in two different cases a=9, c=2 and a=8, c=1. In both cases b doesn't affect the divisibility of abccba by 7, so it can therefore take any value from 19 (that is 7 values in both cases since b cannot equal a or c)
Case 1: a (1 value), b (7 values), and c (1 value) = 9 Case 2: a (1 value), b (7 values), and c (1 value) = 9
Total possible numbers that satisfy the above restrictions: 9+9=18



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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Jan 2020, 05:43
eduardolarrranaga wrote: Can we solve the problem this way?
942 = 3^2 2^2 2^1
Total Factors = (2+1) * (2+1) * (1+1) = 18 942 is given as an example to show what the question means. The number of factors of 942 has nothing to do with the solution.
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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21 Jan 2020, 07:00
The difference of a 3digit number and its reverse is always divisible by 99. Now, 99 does not have a factor of 7 in it. Expressing the 3digit number as xyz and its reverse as zyx, we have their difference to be 99 (xz) { because xyz can also be written as 100x + 10y + z and zyx as 100z + 10y + x}. So, the (xz) has to have a factor of 7 if the difference has to be a multiple of 7 as the question says. Considering that all the digits involved here are nonzero, (xz) can be a factor of 7 in only two ways i.e. x=9 and z = 2 AND x=8 and z=1. For each of these cases, y can be dealt in 9 ways i.e. y can be any digit from 1 to 9. Therefore, we have a total of 2*9 = 18 numbers. The correct answer option is E. Hope that helps!
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a
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